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rewona [7]
3 years ago
15

A fire that is burning wood will release water vapor and carbon dioxide true or false?

Chemistry
1 answer:
Triss [41]3 years ago
6 0

Answer:

true.

Explanation:

You might be interested in
We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh
k0ka [10]

Answer:

The  ideal molar volume is  \frac{V}{n}  =V_z=  0.001095 \ m^3/mol  

The  Z factor is  Z = 0.09997

The  real molar volume is \frac{V_r}{n} = V_k=   0.0001095\ \frac{m^3}{mol}

Explanation:

From the question we are told that

    The pressure is  P  = 23 \ bar =  23 *10^5 Pa

      The temperature is  T  =  30 ^ oC  = 303 \ K

According to the ideal gas equation we have that

          PV  =  nRT

=>      \frac{V}{n}=V_z= \frac{RT}{P}

Where  \frac{V}{n } is the molar volume  and  R is the gas constant with value

            R  =  8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}

substituting values

            \frac{V}{n}  =V_z=  \frac{ 8.314 *  303}{23 *10^{5}}

             \frac{V}{n}  =V_z=  0.001095 \ m^3/mol            

The  compressibility factor of the gas is mathematically represented  as

            Z = \frac{P *  V_z}{RT}

substituting values        

          Z = \frac{23 *10^{5} *   0.001095}{8.314 * 303}

          Z = 0.09997

Now the real molar volume is evaluated as

         \frac{V_r}{n} = V_k=  \frac{Z *  RT }{P}

substituting values

             \frac{V_r}{n} = V_k=   \frac{0.09997 *  8.314 *  303}{23 *10^{5}}

             \frac{V_r}{n} = V_k=   0.0001095\ \frac{m^3}{mol}

8 0
3 years ago
A sample of a compound that contains only the elements C, H, and N is completely burned in O₂ to produce 44.0 g of CO₂, 45.0 g o
koban [17]

Answer:

CH₅N

Explanation:

In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:

(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C

Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:

(45.0 g)(mol/18.02g) = 2.497...mol H₂O

Moles of H is found using the molar ratio of 2H:1H₂O:

(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H

The ratio of H to C in the compound is:

(4.994...mol H)/(0.99977... mol C) = 5 H:C

Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.

3 0
2 years ago
What is the speed at which molecules or atoms move dependent on temperature and state of matter.
den301095 [7]
The hotter it gets, the faster molecules move, solid form is in low temperature, liquid in medium temperature and gas in high temperature.
3 0
3 years ago
How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol
netineya [11]

The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

Enthalpy is the amount of system heat at constant pressure.

The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.

\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The standard unit is kilojoules (kJ)

The enthalpy change symbol (ΔH) is usually written behind the reaction equation.

Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.

Examples of water evaporation:

 H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ

The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid

  • 155 g of butane

relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol

tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

\large mole\:=\:\large \frac{155}{58}

mole = 2,672

Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:

23.1 kJ / mol x 2,672 mol = 61,723 kJ

<h3>Learn more</h3>

the heat of vaporization

brainly.com/question/11475740

The latent heat of vaporization

brainly.com/question/10555500

brainly.com/question/4176497

Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic

4 0
3 years ago
Read 2 more answers
SCIENCE QUEASTION I PUTED IT IN CHEMISTRY BECAUSE BRAINLY HAS NO SCIENCE WILL MARK BRAINLIEST
Elena L [17]

#1. The condition of the atmosphere at a certain time and certain place

#2. The rate of evaporation is EQUAL to the rate of condensation

Please mark Brainliest if this was helpful! :)

7 0
3 years ago
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